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Product Description

The idea for this puzzle came from Jerry Slocum and Jack Botermans "New Book of Puzzles".

The object of the puzzle is to get all 4 sides to add to 555, the height of the Washington Monument! Each of the cubes will rotate on the internal post.

A great challenge for the math inclined as it is VERY difficult. It might make a good class project from middle school to high school. There are only 9 possible solutions.

Here is a statistical analysis of the puzzle.

So how many different ways are there to align the 10 blocks?To derive this number we’ll start by orienting the lowest block on the base. Since there are no markings on the base that need to be lined up with, the block can be freely rotated with no constraints. There is no wrong way to orient the first block. It’s a freebee.The second block (second from the bottom) gets oriented next. The second block can be rotated into 4 different positions to have its 4 numbers line up with the 4 numbers on the first block. This results in the first two blocks being able to form 4 combinations. This can also be viewed as 1 x 4; one way to place the first block and 4 ways to place the second block. Once the third block is added, for every combination of the first 2 blocks, the third block can be placed in 4 different rotations. This provides 4 x 4 or 16 combinations for the first 3 blocks. When the fourth block is added, for each of the 16 combinations of the first 3 blocks, the fourth block can be placed in 4 different rotations. This results in 16 x 4 or 64 combinations.

The pattern of multiplying the number of combinations by 4 for each block holds for the remaining blocks resulting in 1 x 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4 x 4 (also written as 4**9) or 262,144 combinations for all 10 blocks.

Note that there is only nine 4’s that get multiplied even though there are 10 blocks. This is because the first block is not bounded. If the base had 4 numbers that the first block had to align with, then there would have been ten 4’s. In general the equation for any amount of blocks and numbers per block would result in the following number of combinations with the first block unconstrained: NumbersPerBlock**(NumberOfBlocks – 1)

If the first block were restrained by adding by adding numbers to the base, the equation would be :numbers per block ** number of blocks

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